The incircle of $\Delta ABC$ touches $BC$, $CA$, and $AB$ at $D$, $E$, and $F$ respectively. $X$ is a point inside $\Delta ABC$ such that the incircle of $\Delta XBC$ touches $BC$ at $D$ also, and touches $CX$ and $XB$ at $Y$ and $Z$, respectively. Prove that $EFZY$ is a cyclic quadrilateral.
In this question, I can follow up to the point $P$, $Z$, $Y$ are declared collinear. But after that , how does $$PF\cdot PE=PZ\cdot PY$$ confirm that $EFZY$ is a cyclic quadrilateral?
In the figure, $PE \cdot PF = PZ \cdot PY \implies \Delta PEY \sim \Delta PZF$
Hence $\angle PEY = \angle PZF $
and $EFZY$ is a cyclic quadrilateral.